Trigonometry Examples

tan (x) = \frac{5}{8}

$\tan (x) = \frac{5}{8}$

Step 1

Take the inverse tangent of both sides of the equation to extract

x

$x$ from inside the tangent.

x = arctan (\frac{5}{8})

$x = \arctan (\frac{5}{8})$

Step 2

Simplify the right side.

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Step 2.1

Evaluate

arctan (\frac{5}{8})

$\arctan (\frac{5}{8})$ .

x = 0.55859931

$x = 0.55859931$

x = 0.55859931

$x = 0.55859931$

Step 3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from

π

$π$ to find the solution in the fourth quadrant.

x = (3.14159265) + 0.55859931

$x = (3.14159265) + 0.55859931$

Step 4

Solve for

x

$x$ .

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Step 4.1

Remove parentheses.

x = 3.14159265 + 0.55859931

$x = 3.14159265 + 0.55859931$

Step 4.2

Remove parentheses.

x = (3.14159265) + 0.55859931

$x = (3.14159265) + 0.55859931$

Step 4.3

Add

3.14159265

$3.14159265$ and

0.55859931

$0.55859931$ .

x = 3.70019196

$x = 3.70019196$

x = 3.70019196

$x = 3.70019196$

Step 5

Find the period of

tan (x)

$\tan (x)$ .

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Step 5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 5.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

1

$1$ is

1

$1$ .

\frac{π}{1}

$\frac{π}{1}$

Step 5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 6

The period of the

tan (x)

$\tan (x)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

x = 0.55859931 + π n, 3.70019196 + π n

$x = 0.55859931 + π n, 3.70019196 + π n$ , for any integer

n

$n$

Step 7

Consolidate

0.55859931 + π n

$0.55859931 + π n$ and

$3.70019196 + π n$ to

$0.55859931 + π n$ .

$x = 0.55859931 + π n$ , for any integer

$n$